Question

Cho \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)

So sánh a + b với 2c

Answer 1

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

<=>\(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

<=>\(2\left(a^2+b^2+c^2-ab-bc-ca\right)\)\(=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

<=>\(0=4\left(a^2+b^2+c^2-ab-bc-ca\right)-\)\(2\left(a^2+b^2+c^2-ab-bc-ca\right)\)

<=>\(2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Dấu "=" xảy ra khi \(\left(a-b\right)^2=\left(b-c\right)^2=\left(c-a\right)^2=0\)

<=>a-b=b-c=c-a<=>a=b=c<=>a+b=2c